NameError when appending to a List inside a function (Python)
-4
Why do I get the error NameError: name 'l1' is not defined when I run the code below, will placing the l1 and l2 empty lists outside the functions allow me to append/use the return statements?
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
call = function()
print(l1)
print(l2)
python list function
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
add a comment |
-4
Why do I get the error NameError: name 'l1' is not defined when I run the code below, will placing the l1 and l2 empty lists outside the functions allow me to append/use the return statements?
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
call = function()
print(l1)
print(l2)
python list function
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
2
l1andl2are not defined outside of the function (something likel1=something).callon the other hand, is a tuple containing the two results from your functions. Perhaps you want to change it tol1, l2 = function(), or leaving as is, and insteadprint(call[0])andprint(call[1]).
– kabanus
Nov 14 '18 at 7:58
Python Scopes and Namespaces
– Christian König
Nov 14 '18 at 8:04
add a comment |
-4
-4
-4
Why do I get the error NameError: name 'l1' is not defined when I run the code below, will placing the l1 and l2 empty lists outside the functions allow me to append/use the return statements?
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
call = function()
print(l1)
print(l2)
python list function
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
Why do I get the error NameError: name 'l1' is not defined when I run the code below, will placing the l1 and l2 empty lists outside the functions allow me to append/use the return statements?
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
call = function()
print(l1)
print(l2)
python list function
python list function
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
asked Nov 14 '18 at 7:56
CosmicCatCosmicCat
825
825
2
l1andl2are not defined outside of the function (something likel1=something).callon the other hand, is a tuple containing the two results from your functions. Perhaps you want to change it tol1, l2 = function(), or leaving as is, and insteadprint(call[0])andprint(call[1]).
– kabanus
Nov 14 '18 at 7:58
Python Scopes and Namespaces
– Christian König
Nov 14 '18 at 8:04
add a comment |
2
l1andl2are not defined outside of the function (something likel1=something).callon the other hand, is a tuple containing the two results from your functions. Perhaps you want to change it tol1, l2 = function(), or leaving as is, and insteadprint(call[0])andprint(call[1]).
– kabanus
Nov 14 '18 at 7:58
Python Scopes and Namespaces
– Christian König
Nov 14 '18 at 8:04
2
2
l1 and l2 are not defined outside of the function (something like l1=something). call on the other hand, is a tuple containing the two results from your functions. Perhaps you want to change it to l1, l2 = function(), or leaving as is, and instead print(call[0]) and print(call[1]).– kabanus
Nov 14 '18 at 7:58
l1 and l2 are not defined outside of the function (something like l1=something). call on the other hand, is a tuple containing the two results from your functions. Perhaps you want to change it to l1, l2 = function(), or leaving as is, and instead print(call[0]) and print(call[1]).– kabanus
Nov 14 '18 at 7:58
Python Scopes and Namespaces
– Christian König
Nov 14 '18 at 8:04
Python Scopes and Namespaces
– Christian König
Nov 14 '18 at 8:04
add a comment |
1 Answer
1
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0
You can't call it like that, l1 and l2 are undefined then, so need to do:
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
l1,l2 = function()
print(l1)
print(l2)
Output:
[1]
[2]
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
add a comment |
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active
oldest
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oldest
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oldest
votes
0
You can't call it like that, l1 and l2 are undefined then, so need to do:
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
l1,l2 = function()
print(l1)
print(l2)
Output:
[1]
[2]
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
add a comment |
0
You can't call it like that, l1 and l2 are undefined then, so need to do:
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
l1,l2 = function()
print(l1)
print(l2)
Output:
[1]
[2]
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
add a comment |
0
0
0
You can't call it like that, l1 and l2 are undefined then, so need to do:
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
l1,l2 = function()
print(l1)
print(l2)
Output:
[1]
[2]
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
You can't call it like that, l1 and l2 are undefined then, so need to do:
def function():
l1 =
l2 =
for x in range(1):
if 3 > 2:
l1.append(1)
l2.append(2)
return l1, l2
l1,l2 = function()
print(l1)
print(l2)
Output:
[1]
[2]
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
answered Nov 14 '18 at 8:01
U9-ForwardU9-Forward
16k51543
16k51543
add a comment |
add a comment |
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2
l1andl2are not defined outside of the function (something likel1=something).callon the other hand, is a tuple containing the two results from your functions. Perhaps you want to change it tol1, l2 = function(), or leaving as is, and insteadprint(call[0])andprint(call[1]).– kabanus
Nov 14 '18 at 7:58
Python Scopes and Namespaces
– Christian König
Nov 14 '18 at 8:04